Proving irreducibility of polynomial in (Z/2Z)[X]

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Stochastic_Jimmy

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If I want to prove that the polynomial \(\displaystyle f(x) = x^8 + x^4 + x^3 +x + 1 \) is irreducible in \(\displaystyle (\mathbb{Z}/2\mathbb{Z})[X] \), is there a standard way to proceed with a proof of this type?

One method I was thinking about is the following: I think that if the polynomial \(\displaystyle f \), which is of degree 8, is actually reducible then it would have to an irreducible factor of degree 1 or 2 or 3 or 4. And since there aren't many irreducible polynomials of those degrees in \(\displaystyle (\mathbb{Z}/2\mathbb{Z})[X] \) I could show that none of them are factors of \(\displaystyle f \).

But I'm not sure if this is correct, and even if it is correct it seems a bit crude.

Thank you for any comments. I really appreciate it.
 
In case you saw it, my first reply (that I deleted) was me having misread one of powers of x (change the x to an x^2 and it becomes factorable). Your approach is fine.

No factor of degree 1 => No factor of degree 7.
No factor of degree 2 => No factor of degree 5. etc

edit: I spoke too soon!

Add "0" to your polynomial in the following way: 2(x^7+x^5+x^4+x^3+x^2)

You end up with:

\(\displaystyle x^8+2x^7+2x^5+3x^4+2x^3+2x^2+x+1\)

rearrange as follows:

\(\displaystyle x^8+x^7+x^5+x^4+x^3 + x^7+x^6+x^4+x^3+x^2 + x^5+x^4+x^2+x+1 \)

Then apply group factoring to every 5 terms and you will see it is a composite polynomial
 
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Thanks a lot for the response. Are you sure that the polynomial
daon2 said:
rearrange as follows:

\(\displaystyle x^8+x^7+x^5+x^4+x^3 + x^7+x^6+x^4+x^3+x^2 + x^5+x^4+x^2+x+1 \)
Click to expand...

is actually the same as the original? I think that mod 2 the polynomial \(\displaystyle x^8+x^7+x^5+x^4+x^3 + x^7+x^6+x^4+x^3+x^2 + x^5+x^4+x^2+x+1 \) actually simplifies to \(\displaystyle x^{8}+{x}^{4}+{x}^{6}+x+1 \) instead of the original \(\displaystyle x^{8}+{x}^{4}+{x}^{3}+x+1 \), but I could definitely be mistaken.
 
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